Problem A. 877. (March 2024)

A. 877. Convex quadrilateral \(\displaystyle ABCD\) is circumscribed about circle \(\displaystyle \omega\). A tangent to \(\displaystyle \omega\) parallel to diagonal \(\displaystyle AC\) meets diagonal \(\displaystyle BD\) at point \(\displaystyle P\) outside of \(\displaystyle \omega\). The second tangent from \(\displaystyle P\) to \(\displaystyle \omega\) touches \(\displaystyle \omega\) at point \(\displaystyle T\). Prove that \(\displaystyle \omega\) and the circumcircle of triangle \(\displaystyle ATC\) are tangent.

Proposed by Nikolai Beluhov, Bulgaria

(7 pont)

Deadline expired on April 10, 2024.

Lemma 1. Let \(\displaystyle A'\) be a point on line \(\displaystyle AC\) outside of \(\displaystyle \omega\), let the two tangents from \(\displaystyle A'\) to \(\displaystyle \omega\) meet line \(\displaystyle BD\) at points \(\displaystyle B'\) and \(\displaystyle D'\), and let the second tangents from \(\displaystyle B'\) and \(\displaystyle D'\) to \(\displaystyle \omega\) meet at point \(\displaystyle C'\). Then \(\displaystyle C'\) lies on line \(\displaystyle AC\).

(Note that we allow each one of \(\displaystyle A'\), \(\displaystyle B'\), \(\displaystyle C'\), and \(\displaystyle D'\) to be a point at infinity. Furthermore, when either one of points \(\displaystyle B'\) and \(\displaystyle D'\) lies on \(\displaystyle \omega\), the ``second'' tangent from it to \(\displaystyle \omega\) is the unique tangent at it to \(\displaystyle \omega\).)

Proof. In a circumscribed quadrilateral, the pole of each diagonal with respect to the incircle lies on the other diagonal.

We apply this to quadrilaterals \(\displaystyle ABCD\) and \(\displaystyle A'B'C'D'\). Since lines \(\displaystyle BD\) and \(\displaystyle B'D'\) coincide, it follows that lines \(\displaystyle A'C'\) and \(\displaystyle AC\) meet both at \(\displaystyle A'\) and at the pole \(\displaystyle V\) of line \(\displaystyle BD \equiv B'D'\) with respect to \(\displaystyle \omega\). Thus lines \(\displaystyle AC\) and \(\displaystyle A'C'\) coincide unless \(\displaystyle A' \equiv V\). In the latter case, we let \(\displaystyle A'\) approach \(\displaystyle V\) and we take the limit. \(\displaystyle \square\)

Lemma 2. The correspondence \(\displaystyle A' \leftrightarrow C'\) is projective.

Proof. Let lines \(\displaystyle A'B'\), \(\displaystyle B'C'\), \(\displaystyle C'D'\), and \(\displaystyle D'A'\) touch \(\displaystyle \omega\) at points \(\displaystyle K\), \(\displaystyle L\), \(\displaystyle M\), and \(\displaystyle N\), respectively.

Let lines \(\displaystyle AC\) and \(\displaystyle BD\) meet at \(\displaystyle E\). Then also \(\displaystyle E = KM \cap LN\) since quadrilateral \(\displaystyle A'B'C'D'\) is circumscribed about \(\displaystyle \omega\).

Let line \(\displaystyle AC\) meet \(\displaystyle \omega\) at points \(\displaystyle X\) and \(\displaystyle Y\). Then

\(\displaystyle (A'XEY) \stackrel{K}{=} (KXMY) \stackrel{E}{=} (MYKX) \stackrel{M}{=} (C'YEX).\)

(Note that the middle transformation is not projection through \(\displaystyle E\) but inversion of center \(\displaystyle E\).) \(\displaystyle \square\)

To complete the solution, let line \(\displaystyle PT\) meet line \(\displaystyle AC\) at point \(\displaystyle Z\) and let \(\displaystyle f\) be the correspondence of Lemma 2. Then \(\displaystyle f\) swaps the pairs \(\displaystyle A \leftrightarrow C\) and \(\displaystyle Z \leftrightarrow \infty\). So does inversion of center \(\displaystyle Z\) and a suitable radius. Thus \(\displaystyle f\) coincides with that inversion.

Since the pole of line \(\displaystyle AC\) with respect to \(\displaystyle \omega\) lies on line \(\displaystyle BD\), by a limiting argument we see that \(\displaystyle f\) swaps also the pair \(\displaystyle X \leftrightarrow Y\). (Where \(\displaystyle X\) and \(\displaystyle Y\) are defined as in the proof of Lemma 2.) Therefore, \(\displaystyle ZT^2 = ZX \cdot ZY = ZA \cdot ZC\), and so the circumcircle of triangle \(\displaystyle ATC\) is tangent to line \(\displaystyle ZT\) at \(\displaystyle T\), as needed.

Remark 1. Lemma 1 shows that a circumscribed quadrilateral can ``flex'' so that its incircle and the lines of its diagonals remain fixed.

We can also phrase this as a closure theorem. Consider a circle \(\displaystyle \omega\) and two lines \(\displaystyle \ell\) and \(\displaystyle m\). Take a point \(\displaystyle P_1\) on \(\displaystyle \ell\) outside of \(\displaystyle \omega\) and let one of the two tangents from \(\displaystyle P_1\) to \(\displaystyle \omega\) meet \(\displaystyle m\) at point \(\displaystyle P_2\). Next let the second tangent from \(\displaystyle P_2\) to \(\displaystyle \omega\) meet \(\displaystyle \ell\) at point \(\displaystyle P_3\), and so on and so forth. If \(\displaystyle P_5 \equiv P_1\) for some choice of \(\displaystyle P_1\), then \(\displaystyle P_5 \equiv P_1\) for all choices of \(\displaystyle P_1\). Furthermore, the chain closes in this way if and only if the pole of each one of lines \(\displaystyle \ell\) and \(\displaystyle m\) with respect to \(\displaystyle \omega\) lies on the other one.

Remark 2. One special case interesting in its own right is when \(\displaystyle \angle ADC\) is straight. With the points renamed to fit this setting somewhat better, we obtain the following result.

Corollary. Let the incircle \(\displaystyle \omega\) of triangle \(\displaystyle ABC\) touch side \(\displaystyle BC\) at point \(\displaystyle D\). Let the second tangent to \(\displaystyle \omega\) parallel to line \(\displaystyle BC\) meet segment \(\displaystyle AD\) at point \(\displaystyle P\) and let the second tangent from \(\displaystyle P\) to \(\displaystyle \omega\) touch \(\displaystyle \omega\) at point \(\displaystyle T\). Then \(\displaystyle \omega\) and the circumcircle of triangle \(\displaystyle BTC\) are tangent.

This admits an elementary proof independent from our solution to the problem. Here is a quick sketch: Let \(\displaystyle r\) be the radius of \(\displaystyle \omega\), let the second tangent to \(\displaystyle \omega\) parallel to line \(\displaystyle BC\) meet sides \(\displaystyle AB\) and \(\displaystyle AC\) at points \(\displaystyle K\) and \(\displaystyle L\), respectively, let the same tangent touch \(\displaystyle \omega\) at point \(\displaystyle Q\), and, finally, let line \(\displaystyle PT\) meet line \(\displaystyle BC\) at point \(\displaystyle X\). Then \(\displaystyle KP = LQ\), \(\displaystyle KQ = LP\), \(\displaystyle BD = r^2/KQ\), \(\displaystyle CD = r^2/LQ\), and \(\displaystyle XD = r^2/PQ\). These identities suffice to verify that \(\displaystyle XB \cdot XC = XD^2\).

Statistics:

6 students sent a solution.
7 points:	Bodor Mátyás, Forrai Boldizsár, Philip Stefanov, Wiener Anna.
6 points:	Varga Boldizsár.
2 points:	1 student.

Problems in Mathematics of KöMaL, March 2024